Problem 53 - Chemistry Matter and Change

Question

A photon has an energy of \(1.10 \times 10^{-13} \mathrm{J.}\) What is the photon's wavelength? What type of electromagnetic radiation is it?

Step-by-Step Solution

Answer

The wavelength is approximately \(1.807 \times 10^{-12}\) meters, classifying it as X-ray radiation.

Step 1: Understand the Problem

We need to find the wavelength of a photon given its energy, and then determine the type of electromagnetic radiation it corresponds to.

Step 2: Apply the Energy-Wavelength Relationship

Use the formula for energy of a photon: \[ E = \frac{hc}{\lambda} \]where \( E \) is the energy, \( h \) is Planck's constant \( \left(6.626 \times 10^{-34} \text{J} \cdot \text{s} \right) \), \( c \) is the speed of light \( \left(3 \times 10^8 \text{m/s} \right) \), and \( \lambda \) is the wavelength. Rearrange to find the wavelength:\[ \lambda = \frac{hc}{E} \]

Step 3: Substitute the Known Values

Substitute the given energy \( E = 1.10 \times 10^{-13} \text{ J} \) into the equation:\[ \lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.10 \times 10^{-13}} \]

Step 4: Perform the Calculation

Calculate the expression:\[ \lambda = \frac{1.9878 \times 10^{-25}}{1.10 \times 10^{-13}} \approx 1.807 \times 10^{-12} \text{ m} \]

Step 5: Identify the Radiation Type

Compare the wavelength of \(1.807 \times 10^{-12} \text{ m}\) with the electromagnetic spectrum. This wavelength falls in the range of X-rays, which is typically from about \(10^{-11}\) to \(10^{-8}\) meters.

Key Concepts

Wavelength Calculation

The wavelength of a photon is intricately linked to its energy. This relationship is crucial for understanding how we can measure and identify different types of electromagnetic radiation. To find the wavelength when a photon's energy is known, we rely on the formula: \[ \lambda = \frac{hc}{E} \] Here, \( \lambda \) signifies the wavelength, \( h \) is Planck's constant, \( c \) stands for the speed of light, and \( E \) represents energy. By rearranging this formula, we can isolate \( \lambda \) and substitute the known values to calculate the wavelength. This allows us to precisely determine the type of radiation, whether it's visible light or part of another spectrum, such as X-rays.

Electromagnetic Spectrum

The electromagnetic spectrum describes the range of all possible frequencies of electromagnetic radiation. It spans from very low-energy radio waves to high-energy gamma rays. Each segment of the spectrum has its specific characteristics:

• Radio waves are at the low-energy end and are used for communication technologies.
• Visible light is a small portion that humans can see.
• X-rays are higher in energy and are used in medical imaging.

Understanding this spectrum is essential because it helps us classify radiation based on its wavelength or frequency. For instance, identifying a wavelength like \(1.807 \times 10^{-12} \text{ m}\) as X-rays helps in utilizing this property effectively in practical applications.

Planck's Constant

Planck's constant is a fundamental physical constant, represented as \( h \), with a value of approximately \( 6.626 \times 10^{-34} \text{J} \cdot \text{s} \). This constant plays a vital role in quantum mechanics and the calculation of photon energies. It serves as the bridge linking a photon's energy to its frequency, precisely by the equation: \[ E = hf \] Furthermore, Planck's constant is an indispensable part of the energy-wavelength relation \( E = \frac{hc}{\lambda} \). Understanding this constant enables deeper insights into the quantized nature of light and energy – a cornerstone of modern physics.

X-rays

X-rays are a form of electromagnetic radiation with wavelengths ranging from about \(10^{-11}\) to \(10^{-8}\) meters. This places them at the shorter wavelength, high-energy end of the electromagnetic spectrum. X-rays are best known for their use in medical imaging to view the inside of the body, evaluate broken bones, and even detect cancer. Their high energy allows them to pass through various materials, making them exceptionally useful for seeing through objects that visible light cannot penetrate. Moreover, understanding the precise wavelength—like the one calculated, \(1.807 \times 10^{-12} \text{ m}\)—is crucial for classifying electromagnetic radiation and designing tools and procedures.